pub fn new_birthday_probability(n: u32) -> f64 {
    if n < 2 {
        return 0.0;
    }
    
    // 1. 计算没有重复生日的概率
    let mut no_match_prob = 1.0;
    for i in 1..n {
        no_match_prob *= (365 - i) as f64 / 365.0;
    }
    
    // 2. 计算有重复生日的概率
    let result = 1.0 - no_match_prob;
    (result * 10000.0).round() / 10000.0
}
